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Group Theory and the Rubik's Cube

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Rubiks Cube Group Theory

Text of Group Theory and the Rubik's Cube

Group Theory and the Rubiks Cube

Janet Chen

A Note to the ReaderThese notes are based on a 2-week course that I taught for high school students at the Texas State Honors Summer Math Camp. All of the students in my class had taken elementary number theory at the camp, so I have assumed in these notes that readers are familiar with the integers mod n as well as the units mod n. Because one goal of this class was a complete understanding of the Rubiks cube, I have tried to use notation that makes discussing the Rubiks cube as easy as possible. For example, I have chosen to use right group actions rather than left group actions.


IntroductionHere is some notation that will be used throughout.

Z N Q R Z/nZ (Z/nZ)

the the the the the the

set set set set set set

of of of of of of

integers . . . , 3, 2, 1, 0, 1, 2, 3, . . . positive integers 1, 2, 3, . . . rational numbers (fractions) real numbers integers mod n units mod n

The goal of these notes is to give an introduction to the subject of group theory, which is a branch of the mathematical area called algebra (or sometimes abstract algebra). You probably think of algebra as addition, multiplication, solving quadratic equations, and so on. Abstract algebra deals with all of this but, as the name suggests, in a much more abstract way! Rather than looking at a specic operation (like addition) on a specic set (like the set of real numbers, or the set of integers), abstract algebra is algebra done without really specifying what the operation or set is. This may be the rst math youve encountered in which objects other than numbers are really studied! A secondary goal of this class is to solve the Rubiks cube. We will both develop methods for solving the Rubiks cube and prove (using group theory!) that our methods always enable us to solve the cube.

ReferencesDouglas Hofstadter wrote an excellent introduction to the Rubiks cube in the March 1981 issue of Scientic American. There are several books about the Rubiks cube; my favorite is Inside Rubiks Cube and Beyond by Christoph Bandelow. David Singmaster, who developed much of the usual notation for the Rubiks cube, also has a book called Notes on Rubiks Magic Cube, which I have not seen. For an introduction to group theory, I recommend Abstract Algebra by I. N. Herstein. This is a wonderful book with wonderful exercises (and if you are new to group theory, you should do lots of the exercises). If you have some familiarity with group theory and want a good reference book, I recommend Abstract Algebra by David S. Dummit and Richard M. Foote.




To understand the Rubiks cube properly, we rst need to talk about some dierent properties of functions. Denition 1.1. A function or map f from a domain D to a range R (we write f : D R) is a rule which assigns to each element x D a unique element y R. We write f (x) = y . We say that y is the image of x and that x is a preimage of y . Note that an element in D has exactly one image, but an element of R may have 0, 1, or more than 1 preimage. Example 1.2. We can dene a function f : by f (x) = x2 . If x is any real number, its image is the real number x2 . On the other hand, if y is a positive real number, it has two preimages, y and y . The real number 0 has a single preimage, 0; negative numbers have no preimages. y Functions will provide important examples of groups later on; we will also use functions to translate information from one group to another. Denition 1.3. A function f : D R is called one-to-one if x1 = x2 implies f (x1 ) = f (x2 ) for x1 , x2 D. That is, each element of R has at most one preimage. Example 1.4. Consider the function f : dened by f (x) = x + 1. This function is one-to-one since, if x1 = x2 , then x1 + 1 = x2 + 1. If x is an integer, then it has a single preimage (namely, x 1). If x is not an integer, then it has no preimage.




The function f : dened by f (x) = x2 is not one-to-one, since f (1) = f (1) but 1 = 1. Here, 1 has two preimages, 1 and 1. y Denition 1.5. A function f : D R is called onto if, for every y R, there exists x D such that f (x) = y . Equivalently, every element of R has at least one preimage.


Z R dened by f (x) = x + 1 is not onto since non-integers do not have Z Z dened by f (x) = x + 1 is onto. The function f : Z Z dened by f (x) = x is not onto because there is no x Z such that f (x) = 2. yExample 1.6. The function f : preimages. However, the function f :2

Exercise 1.7. Can you nd a . . . 1. 2. 3. 4. . . . function . . . function . . . function . . . function which which which which is is is is neither one-to-one nor onto? one-to-one but not onto? onto but not one-to-one? both one-to-one and onto?

Denition 1.8. A function f : D R is called a bijection if it is both one-to-one and onto. Equivalently, every element of R has exactly one preimage. Example 1.9. The function f :

Z Z dened by f (x) = x + 1 is a bijection.


Example 1.10. If S is any set, then we can dene a map f : S S by f (x) = x for all x S . This map is called the identity map, and it is a bijection. y Denition 1.11. If f : S1 S2 and g : S2 S3 , then we can dene a new function f g : S1 S3 by (f g )(x) = g (f (x)). The operation is called composition. Remark 1.12. One usually writes (g f )(x) = g (f (x)) rather than (f g )(x) = f (g (x)). However, as long as we are consistent, the choice does not make a big dierence. We are using this convention because it matches the convention usually used for the Rubiks cube.


Exercises1. Which of the following functions are one-to-one? Which are onto? (a) (b) (c) (d) f f f f : : : :

Z Z dened by f (x) = x + 1. N N dened by f (x) = x + 1. Z Z dened by f (x) = 3x + 1. R R dened by f (x) = 3x + 1.2 2

2. Suppose f1 : S1 S2 and f2 : S2 S3 are one-to-one. Prove that f1 f2 is one-to-one. 3. Suppose f1 : S1 S2 and f2 : S2 S3 are onto. Prove that f1 f2 is onto. 4. Let f1 : S1 S2 , f2 : S2 S3 , and f3 : S3 S4 . Prove that f1 (f2 f3 ) = (f1 f2 ) f3 . 5. Let S be a set. (a) Prove that there exists a function e : S S such that e f = f and f e = f for all bijections f : S S . Prove that e is a bijection. S . (b) Prove that, for every bijection f : S S , there exists a bijection g : S S such that f g = e and g f = e. 6. If f : D R is a bijection and D is a nite set with n elements, prove that R is also a nite set with n elements.




Example 2.1. To get an idea of what groups are all about, lets start by looking at two familiar sets. First, consider the integers mod 4. Remember that /4 is a set with 4 elements: 0, 1, 2, and 3. One of the rst things you learned in modular arithmetic was how to add numbers mod n. Lets write an addition table for /4 .

Z Z1 1 2 3 0


+ 0 1 2 3

0 0 1 2 3

2 2 3 0 1

3 3 0 1 2

Now, were going to rewrite the addition table in a to use the symbol instead of + for addition, and addition table looks like e e e a a b b c c

way that might seem pretty pointless; were just going well write e = 0, a = 1, b = 2, and c = 3. Then, our a b c a b c b c e c e a e a b

Lets do the same thing for ( /5 ) , the set of units mod 5. The units mod 5 are 1, 2, 3, and 4. If you add two units, you dont necessarily get another unit; for example, 1 + 4 = 0, and 0 is not a unit. However, if you multiply two units, you always get a unit. So, we can write down a multiplication table for ( /5 ) . Here it is: 1 2 4 3 1 1 2 4 3 2 2 4 3 1 4 4 3 1 2 3 3 1 2 4



Again, were going to rewrite this using new symbols. Let mean multiplication, and let e = 1, a = 2, b = 4, and c = 3. Then, the multiplication table for ( /5 ) looks like


e a b c e e a b c a a b c e b b c e a c c e a b Notice that this is exactly the same as the table for addition on


Why is it interesting that we get the same tables in these two dierent situations? Well, this enables us to translate algebraic statements about addition of elements of /4 into statements about multiplication of elements of ( /5 ) . For example, the equation x + x = 0 in /4 has two solutions, x = 0 and x = 2. With our alternate set of symbols, this is the same as saying that the equation x x = e has solutions x = e and x = b. If we translate this to ( /5 ) , this says that the solutions of x x = 1 in ( /5 ) are x = 1 and x = 4. That is, 1 and 4 are the square roots of 1 in ( /5 ) , which is exactly right!




In mathematical language, we say that /4 with addition and ( /5 ) with multiplication are isomorphic groups. The word isomorphic means roughly that they have the same algebraic structure; well get into this later. For now, lets just see what a group is. y





Denition 2.2. A group (G, ) consists of a set G and an operation such that: 1. G is closed under . That is, if a, b G, then a b G. Examples:

The set of negative numbers is not closed under multiplication: if we multiply two negative numbers, we get a positive number. 2. is associative. That is, for any a, b, c G, a (b c) = (a b) c. Examples: Addition and multiplication are associative. Subtraction is not associative because a (b c) = (a b) c. 3. There is an identity element e G which satises g = e g = g e for all g G. Examples: For ( (( /5 For ( For (

Z/4Z is closed under +; after all, we wrote down the addition table, which tells us how to add any two elements of Z/4Z and get another element of Z/4Z. Similarly, (Z/5Z) is closed under multiplication. Z is closed under +: if a, b Z, then a + b Z. Similarly, Z is

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